(x + 1) + ( x + 2 ) + . . . . . . . . + (x + 100) = 5750=> x + 1 + x + 2 + x + 3 + . . . . . . .. . .. . . . + x + 100 = 5750=> ( 1 + 2 + 3 + . . . + 100) + ( x + x + x ngoặc này có 100x . . . . . . + x ) = 5750 =>((100+1).100)/2 +100x => 5050+100x=5750 => 100x=700 => x=7
Ta có: (x + 1) + (x + 2) +...+ (x + 100) = 5750
=> (x + x + x +...... +x) + (1 + 2 + 3 + 4 + ... + 100) = 5750
=> 100x + 5050 = 5750
=> 100x = 5750 - 5050
=> 100x = 700
=> x = 700 : 100
=> x = 7